1
$\begingroup$

Define $y_\alpha (t)$ as a solution of the ODE $\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 3y = 3$, with $y_\alpha (0) = \alpha$.

Now I am supposed to argue that if $y_\alpha (0) = \alpha < 0$, then there exist a $\beta >0$, with $y_\alpha (\beta) = 0$.

The general solution of the ODE is given by $y_{gen} (t) = c_1 * e^{-3t} + c_2 * e^{-t} + 1$.

Another question is wheter there are solutions $y(t)$ of the ODE with $t_2 > t_1 >0$, such that $y(t_1) = y(t_2) = 0$. So solutions with more than one zero point.

I get stuck with both questions. Any help is appreciated.

$\endgroup$
1
$\begingroup$

Well, it takes two constants to define a particular solution of a second-order ODE, so your $y_\alpha$ is in fact not a solution, but a whole bunch thereof. Not that it is really important, though.

To answer your first question, see what is the limit of any solution at $t\to\infty$.

To answer the second one... well, why won't you try and build some graphs for some arbitrary values of $c_1,c_2$? It might give you an idea where to look for the desired solution with two zeros.

$\endgroup$
  • $\begingroup$ Since any solution is a lineair combination of the general solution, I guess that limit is equal to 1 if $t \rightarrow \infty$, because of the negative exponents. But what does that say about the existence of a $\beta >0$, with $y_\alpha (\beta) = 0$? $\endgroup$ – clubkli Oct 7 '15 at 18:25
  • $\begingroup$ See, your function at 0 is negative. Since it tends to 1, it must become positive sometime. Since it is continuous, it must reach every intermediate value someplace in between. Now, 0 is just one of such values. $\endgroup$ – Ivan Neretin Oct 7 '15 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.